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Why k in E(k) diagram continuous?

  1. Jul 22, 2009 #1
    i am wondering why k in E(k) diagram is continuous?
    How is it related to discrete reciprocal lattice which are Fourier components?
  2. jcsd
  3. Jul 22, 2009 #2
    It is usually assumed that one is working in the thermodynamic limit, i.e. there is an infinite lattice. In that case, the Fourier domain is compact but continuous.
  4. Jul 22, 2009 #3
    Think about this in 1D. You have an infinite lattice with period a. Say you have some function that exists on the lattice which can be described by a the function sin(k x) where is the position in space and k is some number. Actually you can think of k as an index and the k-space as a list of these indices. So a point in k-space represents a wave in x-space.

    What happens when k is exactly a reciprocal lattice vector, 2 pi/a? Then the sine wave has a period exactly a the period, a.

    What happens when k is smaller than the reciprocal lattice vector? Then the period of the sine wave is longer than a.

    What happens when k is bigger than 2 pi/a? Then the period of the sine wave is smaller than a.

    Hopefully you can see how this is extended into 3D.

    Therefore k-space is a convenient way to describe families of periodic functions that exist in real space. As it happens, we can conveniently describe the solution to Schrodinger's equation for electrons in a crystal using k-space.
  5. Jul 22, 2009 #4
    Dear bpsbps,
    thanks for reply and help,

    Solution of Schrodinger's equation require non zero Fourier components only for reciprocal lattice vector K, which is a discrete set of latice points.
    How this Fourier analysis of Schro eq: and k-space related?

    is any of the continuous points in k-space are Fourier components solution of Schro: eq?

  6. Jul 23, 2009 #5
    A plane wave in real-space is a point in k-space.

    The k-value of the point is 2 pi / lambda where lambda is the real space wavelength of the plane wave.

    Perhaps I do not understand correctly what you are asking.
  7. Jul 23, 2009 #6
    who said it's continuous? It's not continuous unless you have an infinite lattice as was pointed out...

    IN 1-D the spacing between k's is 2pi/L where L is the length of the atomic wire.
  8. Jul 24, 2009 #7
    There are two different things.

    When you have a periodic Hamiltonian, you can block diagonalize it using Bloch's theorem giving the pseudomomentum k as a good quantum number. It turns out the resulting wave functions are periodic in k space with periods equal to the primitive reciprocal lattice vectors. So you can restrict k to live within the first Brillouin zone (the set of points closer to the origin than any other point which is periodic with k = 0).

    Once you have applied Bloch's theorem, you have wavefunctions:
    [tex]\psi_{n,\mathbf{k}}(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}) u_{\mathbf{k}n}(\mathbf{r})[/tex]
    where [tex]u_{\mathbf{k}n}(\mathbf{r})[/tex] is periodic within the unit cell. Since u is periodic in space, Fourier expansion of u will only allow wave vectors which have the same period as u or an integer multiple of the period of u. These wave vectors turn out to be reciprocal lattice vectors.
  9. Jul 24, 2009 #8
    Dear Kanato, Thanks for helpful reply...

    So, to summarize, k space contains number of unit cell (in infinite lattice, it is dense). Due to periodicity of wavefunction in k space, points outside of BZ are taken back to first BZ. In electronic structure calculation, it is done by sampling the first BZ with a fine grid. So this is about e^(ik.r) component of Bloch function.

    second component, namely u_kn(r), is Fourier expansion of solution of Schro Eq. which is infinite discrete lattice points defined by reciprocal lattice vectors. In electronic structure calculation, the approximation is taken by taken large number of Fourier components (as described by planewave cutoff energy).

    My question is the basis of k-space (2*Pi/a) where a is lattice constant is same for both or not.

  10. Jul 24, 2009 #9
    In 1d, the first BZ is:
    [tex] -\pi/a < k < \pi/a[/tex]

    When expanding the wavefunction in a planewave basis:

    [tex] u_{k}(x) = \sum_n c_{kn} \,e^{iG_nx}[/tex]

    [tex] G_n = n \frac{2\pi}a \qquad n \in \{0, \pm 1, \pm 2, ...\} [/tex]

    G_n labels the planewaves. 2pi/a gives the boundaries of k vectors in the first BZ (it's 2pi/a wide). n2pi/a gives you the plane waves you can expand a wavefunction in, where n is an integer.
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